First slide

Theory of expressions

Question

The solution set of ${(x)}^{2} + {(x + 1)}^{2} = 25$ , where $(x)$ is the least integer greater than or equal to $x$ , is

Difficult

A

$(2, 4)$
B

$(- 5, - 4] \cup (2, 3]$
C

$[- 4, - 3) \cup [3, 4)$
D

None of these

Solution

The given equation is ${(x)}^{2} + {(x + 1)}^{2} = 25$
When $x = n \in Z$
Then given equation becomes $\Rightarrow 2 x^{2} + 2 x - 24 = 0$
$\Rightarrow x^{2} + x - 12 = 0$
$\Rightarrow (x + 4) (x - 3) = 0$
$\Rightarrow x = 3, - 4$ .
Thus $x = 3, - 4$ .
When $x = n + k, n \in Z, 0 < k < 1$ ,
then we have, ${(n + 1)}^{2} + {(n + 2)}^{2} = 25$ $\Rightarrow 2 n^{2} + 6 n - 20 = 0$
$\Rightarrow n^{2} + 3 n - 10 = 0$ $\Rightarrow n = 2, - 5$
$∴ x = 2 + k, - 5 + k$ , where $0 < k < 1$
Hence $x \in (- 5, - 4] \cup (2, 3]$ .

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